the rate constant of a reaction …

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the rate constant of a reaction at 500 Kelvin and 700 Kelvin are 0.02 second inverse and 0.07 second respectively calculate the value of A and E

Posted by Kulwant Singh 6 years, 1 month ago

CBSE > Class 12 > Other Subjects

1 answers

Gaurav Seth 6 years, 1 month ago

Relation between equilibrium constant k, E_a and T is given by the equation
ln k = – E_a/RT + ln A
If we have different rate constant at different temperature then given equation can be modified to find out the E_a and A as follows:
log k₂/log k₁ = Ea/2.303 R[T₂-T₁/T₁T₂]...(1)
Putting the values in equation (1) we get,
log 0.07/log 0.02 = E_a/2.303 x 8.314 J K^-1 mol^-1 (700-500/700 x 500)
0.544 = E_a × 5.714 × 10^-4/19.15
E_a = 0.544 × 19.15 / 5.714 × 10^–4 = 18230.8 J
Since k = Ae^-Ea^/RT
Here we can take any value of k and its corresponding temperature
0.02 = Ae^{-18230.8/8.314 × 500}
(Note to calculate e^{-18230.8/8.314 × 500}you can find the antilog of total solved value
antilog of -18230.8/8.314 × 500 is 0.012)
A = 0.02/0.012 = 1.61

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